Is ${347148}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {347148}= &&{3}\cdot100000+ \\&&{4}\cdot10000+ \\&&{7}\cdot1000+ \\&&{1}\cdot100+ \\&&{4}\cdot10+ \\&&{8}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {347148}= &&{3}(99999+1)+ \\&&{4}(9999+1)+ \\&&{7}(999+1)+ \\&&{1}(99+1)+ \\&&{4}(9+1)+ \\&&{8} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {347148}= &&\gray{3\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {3}+{4}+{7}+{1}+{4}+{8} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${347148}$ is divisible by $9$ if ${ 3}+{4}+{7}+{1}+{4}+{8}$ is divisible by $9$ Add the digits of ${347148}$ $ {3}+{4}+{7}+{1}+{4}+{8} = {27} $ If ${27}$ is divisible by $9$ , then ${347148}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${347148}$ must also be divisible by $9$.